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svetoff [14.1K]
3 years ago
8

Which of earths spheres is not included in the phosphorus cycle?

Chemistry
2 answers:
Luda [366]3 years ago
5 0

the answer is B. Atmosphere

UkoKoshka [18]3 years ago
4 0

The phosphorus cycle is the biogeochemical cycle that describes the movement of phosphorus through the lithosphere, hydrosphere, and biosphere.

The atmosphere does not play an important role in the movement of phosphorus, because phosphorus and phosphorus-based compounds are  solids at the symbolic ranges of temperature and pressure present on Earth.

Thus, the earths spheres like biosphere, Cryosphere  and lithosphere are included in the phosphorus cycle but atmosphere is not included in the same.

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Rewrite and Balance the Equation <br> N₂ + H₂ -&gt; NH3
Mars2501 [29]

Answer:N

2

+ 3

H

2

-----> 2N

H

3

Explanation:

N

2

+

H

2

-----> N

H

3

Let us balance this equation by counting the number of atoms on both sides of the arrow.

N

2

+

H

2

-----> N

H

3

N=2 , H=2 N=1, H=3

To balance the number of N atom on Right Hand Side (RHS) , I will add one molecule of N

H

3

on RHS

N

2

+

H

2

-----> 2N

H

3

N=2 , H=2 N=2 , H= 6

To balance the number of H atoms on Left Hand Side (LHS) , I will add two molecules of

H

2

on LHS

N

2

+ 3

H

2

-----> 2N

H

3

N=2 , H=6 N=2 , H= 6

Answer link

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4 0
3 years ago
Read 2 more answers
What were the limition of doberiner classifacation​
alexgriva [62]

Answer: I don’t kno

Explanation:

5 0
3 years ago
When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2
MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is \Delta \ T_f = T_{solvent}- T_{solution}

\Delta \ T_f = 2.9 ^0 \ C

Using the depression in freezing point, the molar depression constant of the solvent K_f = \dfrac{\Delta T_f}{m}

K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}

K_f = 4.82 ^0 C / m}

The freezing point of the solution \Delta T_f = T_{solvent} - T_{solution}

\Delta T_f = 7.3^ 0 \ C

The molality of the solution is:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

Molar depression constant of solvent X, K_f = 4.82 ^0 \ C/m

Hence, using the elevation in boiling point;

the Vant'Hoff factor i = \dfrac{\Delta T_f}{k_f \times m}

i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}

\mathbf {i = 1.51 }

3 0
3 years ago
Which of the following depicts a molecule of DNA? A, B, C, D?
Ghella [55]

Answer:

D. I've watched Jurassic park uwu

7 0
3 years ago
How much heat is added if .0948g of water is increased in temperature by .728 degrees C?
Verdich [7]

Answer:

0.289J of heat are added

Explanation:

We can relate the change in heat of a substance with its increasing in temperature using the equation:

q = m*ΔT*S

<em>Where Q is change in heat</em>

<em>m is mass of substance (In this case, 0.0948g of water)</em>

<em>ΔT = 0.728°C</em>

<em>S is specific heat (For water, 4.184J/g°C)</em>

Replacing:

q = 0.0948g*0.728°C*4.184J/g°C

q = 0.289J of heat are added

5 0
3 years ago
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